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2 years ago

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Two sacks contain the same number of identical apples and are separated by a distance r. The two sacks exert a gravitational for

ce on each other that is opposite in direction but with the same magnitude, F. Half the apples are removed from one sack and placed in the second sack. What is the magnitude of the gravitational force between the two sacks in terms of the original force, F?

1 answer:

stepan [7]2 years ago

3 0

Answer:

Explanation:

im sleppy

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A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object

Darya [45]

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, $F_x=8\ N$

Force acting along +y direction, $F_y=6\ N$

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

$R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}$

Here $\theta=90^{\circ}$

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{8^2+6^2}$

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

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3 years ago

Suppose that instead of being inclined to Earth's orbit around the Sun, the Moonâs orbit was in the same plane as Earthâs orbit

ycow [4]

Answer:

c) 12

Explanation:

A Solar eclipse occurs when The Sun, The Earth and The Moon comes in a straight line with the Moon being in between the Earth and the Sun. At this point the Moon appears to block the Sun and Moon's shadow falls on Earth. This would occur only on the day of the New Moon.

If the Moon's orbit was in the same plane as that of the Earth's orbit. Every new Moon, there would be a Solar Eclipse. The Lunar cycle is of 29.5 Days which means there will be one new Moon every month. So there will be 12 Solar Eclipses every year.

Currently, the orbit of the Moon is tilted at an angle of 5° thus we don't see that many Solar eclipses. Maximum of 5 solar eclipses can occur in an year.

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2 years ago

In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?

IceJOKER [234]

Answer: $a=1016.66 m/s^{2}$

Explanation:

Acceleration $a$ is expressed in the following formula:

$a=\frac{V_{f}-V_{o}}{t}$

Where:

$V_{f}=610 m/s$ is the final velocity of the projectile

$V_{o}=0 m/s$ is the initial velocity of the projectile

$t=0.6 s$ is the time

Solving:

$a=\frac{610 m/s-0 m/s}{0.6 s}$

$a=1016.66 m/s^{2}$ This is the acceleration of the projectile

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3 years ago

An object that is moving must have a change in A) its speed. B) its position. C) its acceleration. D) its applied force.

denis-greek [22]

An object that's moving doesn't necessarily change its speed or acceleration. Also, the force applied to it doesn't need to change ... in fact, a moving object doesn't need ANY force applied to it in order to keep moving.

But any moving object WILL have a change in its position ... THAT's how you know it's moving, and that's WHY you say "It's moving !". (choice-B)

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2 years ago

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

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$N' = \frac{mgcos\theta}{2}$

Pressure is given as the force applied in a given area, that is

$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

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2 years ago